Harrison W. answered • 04/28/20

Yale grad, 100th percentile MCAT, applying to med school

Hi Joselyne. Thanks for your question.

The first thing we need to do is figure out which of the reactants is our limiting reactant. Then we can proceed with the calculations.

In the original balanced equation, we see that for each mole of propane burned, we need 5 moles of oxygen and produce 3 moles of carbon dioxide and 4 moles of water. We are given that the reaction of interest is performed with 3 moles of propane and 10 moles of oxygen. Since we know from the original balanced equation that the ratio of propane to oxygen necessary for the reaction is 1:5 moles, oxygen must be the limiting reagent since with 10 moles of oxygen we will only combust TWO moles of propane (1:5 becomes 2:10 moles). Therefore, we will produce twice what is shown on the reactant side; so 6 moles of carbon dioxide, and 8 moles of water. That gets us almost all the way to our answer - we have 6 moles of CO_{2} produced, and CO_{2} has a molecular weight of (12.011 + 15.999 + 15.999) = 44.009 g/mol. Since we have 6 moles we need to multiply that by 6 to get 264.054 grams of CO_{2} produced.

Harrison W.

Awesome! Thanks for the feedback. Much appreciated04/28/20

Joselyne L.

Thank you for helping! It makes more sense now I appreciate the help.04/28/20